Problem: Solve the equation. $\dfrac{dy}{dx}=\dfrac{x^2}{10y}-\dfrac{2x}{5y}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\pm \sqrt{\dfrac{x^3}{15}-\dfrac{2x^2}{5}+C}$ (Choice B) B $y=\pm \sqrt{\dfrac{x^3}{15}-\dfrac{2x^2}{5}}+C$ (Choice C) C $y=Ce^{^{\frac{10x^3}{3}-20x^2}}$ (Choice D) D $y=\pm e^{^{\frac{10x^3}{3}-20x^2}}+C$
We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{x^2}{10y}-\dfrac{2x}{5y} \\\\ \dfrac{dy}{dx}&=\dfrac{1}{10y}(x^2-4x) \\\\ 10y\,dy&=(x^2-4x)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} 10y\,dy&=(x^2-4x)\,dx \\\\ \int 10y\,dy&=\int (x^2-4x)\,dx \\\\ 5y^2&=\dfrac{x^3}{3}-2x^2+C_1 \\\\ y^2&=\dfrac{x^3}{15}-\dfrac{2x^2}{5}+C \\\\ y&=\pm \sqrt{\dfrac{x^3}{15}-\dfrac{2x^2}{5}+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\pm \sqrt{\dfrac{x^3}{15}-\dfrac{2x^2}{5}+C}$